I am interested in the value of
$$\int_0^\infty e^{-\alpha t}\frac{e^{-\frac{|x-y|^2}{2t}}}{\sqrt{2\pi t}}\, dt $$
this is the laplace transform of the Heat kernel (changing the time variable)
This integral came up in this exercise:
Exercise 3.7. Show that the resolvent for Brownian motion is given by $$U(\alpha)f(x)=\frac{1}{\sqrt{2\alpha}}\int_{-\infty}^\infty f(y)e^{-\sqrt{2\alpha}|x-y|}dy.$$
where
$$U(\alpha)f=\int_0^\infty e^{-\alpha t}T(t)f\,dt,\quad\alpha>0,\tag{3.6}$$
First setps:
\begin{align} U(\alpha)f(x) &= \int_0^\infty e^{-\alpha t} T(t) f (x)\, dt \\ &= \int_0^\infty e^{-\alpha t} \int_{-\infty}^\infty \frac{e^{-\frac{|x-y|^2}{2t}}}{\sqrt{2\pi t}} f (y)\, dy\, dt \\ &= \int_{-\infty}^\infty f(y)\int_0^\infty e^{-\alpha t} \frac{e^{-\frac{|x-y|^2}{2t}}}{\sqrt{2\pi t}} \, dt \, dy \\ \end{align}
So one can guess that
$$ \int_0^\infty e^{-\alpha t} \frac{e^{-\frac{|x-y|^2}{2t}}}{\sqrt{2\pi t}} \, dt = \frac{\exp{- \sqrt{2\alpha}|x-y|}}{\sqrt{2 \alpha}}$$
But How do we get such result?
Here is my attempt at proving this:
$$\int_0^\infty e^{-\alpha t} \frac{e^{-\frac{|x-y|^2}{2t}}}{\sqrt{2\pi t}} \, dt = \int_0^\infty \exp\bigg\{-\alpha t - \frac{|x-y|^2}{2t}\bigg\} \frac{1}{\sqrt{2\pi t}} \, dt \\ = \int_0^\infty \exp\bigg\{-\bigg(\sqrt{\alpha t} - \frac{|x-y|}{\sqrt{2t}}\bigg)^2 + \sqrt{2\alpha}|x-y|\bigg\} \frac{1}{\sqrt{2\pi t}} \, dt\\ = \exp\bigg\{- \sqrt{2\alpha}|x-y|\bigg \}\int_0^\infty \exp\bigg\{-\bigg(\sqrt{\alpha t} - \frac{|x-y|}{\sqrt{2t}}\bigg)^2 \bigg\} \frac{1}{\sqrt{2\pi t}} \, dt$$
So now I am at the point that I need to prove
$$\int_0^\infty \exp\bigg\{-\bigg(\sqrt{\alpha t} - \frac{|x-y|}{\sqrt{2t}}\bigg)^2 \bigg\} \frac{1}{\sqrt{2\pi t}} \, dt = \frac{1}{\sqrt{2 \alpha}} $$
But I don't see how to do it
